On 02/18/2013 07:17 AM, Gottfried Helms wrote:> I've to correct some obvious typing errors:>> Am 18.02.2013 09:38 schrieb David Bernier:>> On 02/18/2013 02:26 AM, Gottfried Helms wrote:>>>>>>>>>> The summatory polynomial for the k'th powers of 1, 2, ... n,>> P(x), has the property that P(x) - P(x-1) = x^k,>> at least for positive integers x.>> I assume k is a positive integer.>>>> So, does there exist a continuous f: [a, oo) such that>> f(x) - f(x-1) = sqrt(x) for any x in [a, oo) ?>>>> Ramanujan wrote a paper on sum of consecutive square roots:>>>> http://oeis.org/A025224/internal>>>> david bernier>>> Hmm, my usual tools give only much diverging series for which my> procedures of divergent summation do not work well if I use -say-> only 64 or 128 terms for the power series.> But I've now converted the problem into one which employs the rationale:>> f(x) = sqrt(1+x^2)> such that f(1) -> sqrt(2) f(f(1)) = sqrt(3) and so on such that we> can write> S(a,b) = sqrt(a) + f(sqrt(a)) + f(f(sqrt(a))) + ... + f...f(sqrt(a))> = f°0(sqrt(a)) + f°1(sqrt(a)) + f°2(sqrt(a)) + ... + f°d(sqrt(a))>> where d=b-a and the circle at f°k(x) indicates the k'th iterate.>> After that the problem can be attacked by the concept of Carleman-matrixes> and their powers. Let F be th carleman-matrix for the function f(x),> then let G = I - F then, if we could invert G in the sense that>> M = G^-1 = I + F + F^2 + F^3 + F^4 + ... (see Neumann-series, wikipedia)>> then we had a solution in terms of a formal power series for>> m(x) = x + f(x) + f°2(x) + f°3(x) + ....>> and m(sqrt(a)) - m(sqrt(b)) would give the required sum-of-sqrt from> sqrt(a) to sqrt(b) in 1-steps progression from its argument a.>> However, G has a zero in the top-left entry(and the whole first column)> and cannot be inverted.> Now there is a proposal which I've seen a couple of times that we invert> simply the upper square submatrix after removing the first (empty) column> in G, let's call this H, and this gives often an -at least- usable> approximate solution, if not arbitrarily exact.>> But again - trying this using Pari/GP leads to nonconclusive results; the> inversion process seems to run in divergent series again.>> Here we have now the possibility to LDU-factorize H into triangular> factors, which each can be inverted, so we have formally>> H = L * D * U> M = H^-1 = U^-1 * (D^-1 * L^-1)>> We cannot perform that multiplication due to still strong divergences> when evaluating the row-column-dotproducts (which is only making explicite> the helpless Pari/GP-attempts for inverting H)>> But we can use the upper triangular U^-1 in the sense of a "change of base"-> operation. Our goal is to have M such that we can write>> (V(sqrt(a)) - V(sqrt(b))) * M[,2] = s(a,b+1) = sqrt(a)+sqrt(a+1)+...+sqrt(b)>> where V(x) means a vandermondevector V(x) = [1,x,x^2,x^3,x^4,....]>> But now we can proceed from the formal formula>> (V(sqrt(a)) - V(sqrt(b))) * M[,2]> = (V(sqrt(a)) - V(sqrt(a))) * U^-1 * ( D^-1 * L^-1)[,2]>> and can compute>> X(sqrt(a),sqrt(a)) = (V(sqrt(a)) - V(sqrt(a))) * U^-1>> exactly to any truncation size because U^-1 is upper triangular and thus> column-finite.>> Then we can as well do>> Q = ( D^-1 * L^-1)[,2]>> which is -besides the truncation to finite size- an exact expression.>> Still we have, that the dot-product>> s(a,b) = X(sqrt(a),sqrt(a)) * Q>> is divergent, but now it seems, that we can apply Euler-summation> for the evaluation of the divergent dot-product.> The results are nice approximations for the first couple of> sums s(1,4), s(2,4) and some more. s(1,9) requires Euler-summation> of some higher order such that I get>> s(1,9) ~ 16.3060> (where 16.3060005260 is exact to the first 11 digits)>> or> s(5,10)= sqrt(5)+sqrt(6)+ ... + sqrt(10)> ~ 16.32199> where 16.3220138163 is exact to the first 11 digits)>>> Don't know how to do better at the moment; surely if the composition> of the entries in the matrices were better known to me, one could> make better, possibly even analytical or at least less diverget, expressions.>> (But I've not enough time to step into this deeper, I'm afraid)

Hi Gottfried,

The exponent for the natural numbers that I'm most interestedin is a = -s , with s = 1/2 + i*t , t real in [0, oo) as:

The other term is one a sum involving the Bernoulli numbers, andand also (1/2) N^(-s) + N^(1-s)/(s-1) [ Edwards].

But a "shortcut" to evaluate:1/1^s +1/2^s + ... + 1/(N-1)^s to high-precision wouldsave time in the Euler-MacLaurin summation for zeta(s),so this connects to algorithms to compute zeta(s) tohigh-precision, a much-studied topic.